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Andrej [43]
3 years ago
9

Lizard able to run in ceilings and walls.why?​

Physics
2 answers:
GarryVolchara [31]3 years ago
8 0

Answer:

because of their web feet

Explanation:

oke

Marysya12 [62]3 years ago
7 0

Answer:

Reptiles have an enormous number of extremely minuscule hairs on the stack of their feet called setae. These little cushions subsequently radically increment the surface zone and come in close contact with the surface on which the reptile is creeping, so the Van der Waals forces kick in.

Explanation:

Hope this helped!

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How many protons, neutrons, and electrons does bi-210 have
nikdorinn [45]

Answer:

Explanation:

The number of neutrons is 127 and the number of protons will be equal to number of electrons. It has 83 protons and a magic number of 127 neutrons. Bi 210 is radioactive isotope of bismuth.

5 0
3 years ago
Determine the value of the force exerted by the surface (normal force) on a
Advocard [28]

Answer:

53.5 N

Explanation:

Vertical component of the F force   50 sin30    = 25 N  upward

   force of gravity = m g = 8 * 9.81 =78.5 N Downward

NET downward force by block on table = net upward force exerted by table =  78.5 -25 =53.5 N

8 0
2 years ago
A grasshopper floating in water generates waves at a rate of three per second with a wavelength of two centimeters. (a) What is
Nastasia [14]

Answer:

(a) 0.33 second

(b) 6 cm/s

Explanation:

Frequency, f = 3 waves per second

wavelength, λ = 2 cm = 0.02 m

(a) The period of wave is defined as the time taken by the wave to complete one oscillation. It is the reciprocal of frequency.

T = 1 / f = 1 / 3 = 0.33 second

(b) the relation between wave velocity, frequency and wavelength is given by

v = f x λ

v = 3 x 0.02 = 0.06 m /s

v = 6 cm /s

8 0
3 years ago
A rocket is fired with an initial VELOCITY OF 100m/s at an angle of 55° above the horizontal, It explodes On the mountain Side 1
GuDViN [60]

Answer

688.32m and 277.44m

Explanation :

⠀

\large{\maltese{\textsf{\underline{To find :-}}}}

The X and Y coordinates of the rocket relative of firing

⠀

⠀

\large{\maltese{\textsf{\underline{Given :-}}}} \\ \\ \sf velocity (v_i) = 100m{s}^{-1} \\ \sf angle ({\theta}_{1}) = 55.0{\degree} \\ \sf time (t) = 12s

⠀

⠀

\Large{\maltese{\textsf{\underline{\underline{Step by Step Solution:-}}}}}

⠀

⠀

<u>The</u><u> </u><u>horizontal</u><u> </u><u>range</u><u> </u><u>of</u><u> </u><u>projectile</u><u> </u><u>at</u><u> </u><u>x</u><u>.</u><u> </u>

⠀

\sf \large{x = v_{xi}} \times t \\ \\ \sf \large{x = v_i \times \cos {\theta}_{i} \times t}

⠀

⠀

\large\textsf{\underline{Now substituting the required values}}

⠀

⠀

\sf x = 300 \times \cos 55{\degree} \times 12 \\ \\ \sf x = 100 \times 0.5756 \times 12 \\ \\ {\underline{\boxed{\bold{ x = 688.32m}}}}

⠀

⠀

The vertical position of projectile at y.

⠀

⠀

\sf \large  y = v_{yi} \times t -  (\frac{1}{2}  \times g  \times {t}^{2}) \\  \\   \sf  \large y = v_i \times  \cos \theta  \times t -  \frac{1}{2} g {t}^{2}

⠀

⠀

\textsf{ \large {\underline{Now substituting the required values}}  }

⠀

⠀

\sf y = 100 \times  \cos55{ \degree} \times 12 -  \frac{1}{2}   \times 9.80 \times  {12}^{2} \\  \\  \sf  y = 100 \times 0.8192 \times 12 - 0.5 \times 9.8 \times 144 \\  \\  \sf y = 983.04 - 705.6 \\  \\  \underline{ \boxed{ \bold{y = 277.44m}}}

⠀

⠀

⠀

<h3><u>Henceforth</u><u>,</u><u> </u><u>the</u><u> </u><u>distance</u><u> </u><u>at</u><u> </u><u>horizon</u><u> </u><u>is</u><u> </u><u>6</u><u>8</u><u>8</u><u>.</u><u>3</u><u>2</u><u>m</u><u> </u><u>and</u><u> </u><u>at</u><u> </u><u>vertical</u><u> </u><u>is</u><u> </u><u>2</u><u>7</u><u>7</u><u>.</u><u>4</u><u>4</u><u>m</u><u>.</u></h3>

8 0
2 years ago
A power plant generates 150 MW of electrical power. It uses a supply of 1000 MW from a geothermal source and rejects energy to t
marissa [1.9K]

Answer:

- the power to the air is 850 MW

- mass flow rate of the air is 84577.11 kg/s

Explanation:

Given the data in the question;

Net power generated; W_{net = 150 MW

Heat input; Q_k = 1000 MW

Power to air = ?

For closed cycles

Power to air Q₀ = Heat input; Q_k - Net power generated; W_{net

we substitute

Power to air Q₀  = 1000 - 150

Q₀ = 850 MW

Therefore,  the power to the air is 850 MW

given that ΔT = 10 °C

mass flow rate of air required will be;

⇒ Q₀ / CpΔT

we know that specific heat of air at p=c ; Cp = 1.005 kJ/kg.K

we substitute

⇒ ( 850 × 10³ ) / [ 1.005 × 10 ]

⇒ ( 850 × 10³ ) / 10.05

⇒ 84577.11 kg/s

Therefore, mass flow rate of the air is 84577.11 kg/s

3 0
3 years ago
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